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Multiple Choice
When 1.045 g of Na2O is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 54.4 °C. Assuming that the specific heat of the solution is 4.18 J/(g·°C) and that the calorimeter itself absorbs a negligible amount of heat, calculate the heat absorbed by the solution.
A
2.09 kJ
B
4.18 kJ
C
8.36 kJ
D
6.12 kJ
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Verified step by step guidance
1
First, identify the mass of the water in the calorimeter. Since the density of water is approximately 1 g/mL, the mass of 50.0 mL of water is 50.0 g.
Next, calculate the change in temperature (ΔT) of the water. This is the final temperature minus the initial temperature: ΔT = 54.4 °C - 25.0 °C.
Use the formula for heat absorbed, q = m × c × ΔT, where m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature. Here, m = 50.0 g, c = 4.18 J/(g·°C), and ΔT is the temperature change calculated in the previous step.
Substitute the values into the formula: q = 50.0 g × 4.18 J/(g·°C) × ΔT. This will give you the heat absorbed by the solution in joules.
Convert the heat absorbed from joules to kilojoules by dividing the result by 1000, since 1 kJ = 1000 J.